Lie group action on manifold and the vector field generated Suppose $G$ is a Lie group and acts smoothly on a compact oriented manifold $M$, and $X\in\mathfrak{g}$ is an element in Lie algebra of $G$ such that the vector field $\tilde X$ generated by $X$
$$
\tilde X(p)=\frac{d}{dt}\exp(tX)\big\vert^{t=0}.p
$$
has isolated zeros. Then $\tilde X$ act on space of vector field on $M$ by Lie bracket: $F\mapsto [\tilde X,F]$. This action induces a well-defined action on $T_pM$ for any point $p$ with $\tilde X(p)=0$. This action is defined by
$
L(X,p):T_pM\to T_pM
$
$
~~~~~~~~~~~~~~~~~~~~~~v\mapsto [\tilde X,F_v](p)~~~~\text{where } F_v\text{ is any vector field with }F_v(p)=v
$
The question is how to show that $L(X,p)$ is invertible.
I tried to show that if there is a non-zero vector $\xi\in T_pM$ s.t. $L(X,p)(\xi)=0$ then $\exp(tX)$ fixes $\exp_p(sv)$ for $t,s$ small. This would then contradict the fact that $p$ is isolated. But I don't know how to show this.
 A: For reasons of laziness/neatness/misuse/abuse of notation, write not only $X$ for the element of the lie algebra $\mathfrak g$ of $G$, but also for the vector field on $M$.
First of all, what exactly is meant by the zeros are 'isolated'? (I am sure there is an accepted definition, but...)
Namely:
If 'isolated' means that there is an open set $U\subset M$ around $p$, on which $X$ vanishes only at $p$, I think the statement is false.
Counter-example (with this definition): on (the non-compact) $M=\mathbb R$, if $(X f) (x) = x^2 f'(x)$,  then $0$ is an 'isolated' zero of $X$. On the other hand,  $[X,{d\over dx}]$  vanishes at $0$. The one-parameter group action on $M$ is $$t\cdot x = {x \over 1 - xt}.$$
To get a counter-example in the compact case, take $M = S^1$ (circle), and set
$$(X f) (\theta) = \sin^2 (\theta) f'(\theta).$$
The action of the one-parameter group is a bit more of a pain to write out ( but without singularities, as there was for the non-compact real-line, which is why I am bothering to state it), but the two examples are basically identical...
On the other hand:
Choose a trivialization chart around $p$ such that we can write
$$X =  y_1(x) {\partial \over \partial x_1} +\cdots + y_n(x) {\partial \over \partial x_n}. $$
If the statement
$$\text{$0$ is an isolated zero of $X$}$$
is equivalent to
$$\text{$y_k$ vanish at $0$, and span (give rise to a basis of) $\mathfrak m/\mathfrak m^2$},$$
where $\mathfrak m$ is the germ of functions vanishing at $0$, (in other words, the 'hypersurfaces' $y_k = 0$ meet transversally at $0$), then the result holds even without compactness of $M$. We can identify $L(X,0)$ with the linear map:
$$ v \mapsto -  \big( v(y_1), \cdots, v(y_n) \big)  .$$
Since [under my (!) hypothesis] $y_k \pmod {\mathfrak m^2}$ form a basis for the dual to the tangent space at $0$, the (original) map is an isomorphism.
